The Ba(OH)2 dissociates as Ba+2 + 2 OH-. H2SO4 dissociates as 2 H+ + SO4-2. When the conductivity is at a minimum, what must be true about the amount of Ba(OH)2 compared to H2SO4? Also, why does it not conduct at this low point?

Question

The Ba(OH)2 dissociates as Ba+2 + 2 OH-. H2SO4 dissociates as 2 H+ + SO4-2.  When the conductivity is at a minimum, what must be true about the amount of Ba(OH)2 compared to H2SO4?  Also, why does it not conduct at this low point?

abb0t · Answer

$H_2SO_4 + Ba(OH)_2 \rightarrow Ba_SO_4 + H_2O$ 
When the conductivity is at a minimum, that means that stoichiometric amounts of $Ba(OH)_2$ and $H2SO4$ are present (neither in excess) and the only materials in the reaction vessel are water and $BaSO_4$ and $ BaSO_4$ is not very soluble and water doesn't ionize appreciably. 
$Ba(OH)_2$ is a strong electrolyte and conducts electrical current easily. 
However, because $BaSO_4$ is formed, it is essentially insoluble, therefore, no ions and no conduction

anonymous · Answer

@abb0t Thanks so much!!